Question

[Nguyễn Duy Long]

1. \(\sqrt{x^2-\frac{7}{x^2}}+\sqrt{x-\frac{7}{x^2}}=x\)

2.\(6\sqrt{2x+4}-4\sqrt{x+1}=x+8\)

3.\(10\sqrt{2x+2}+2\sqrt{x+3}=x+23\)

Answer 1

[Thắng Nguyễn]

a)\(\sqrt{x^2-\frac{7}{x^2}}+\sqrt{x-\frac{7}{x^2}}=x\)

\(\Leftrightarrow\sqrt{x^2-\frac{7}{x^2}}-\frac{3}{2}+\sqrt{x-\frac{7}{x^2}}-\frac{1}{2}-x+2=0\)

\(\Leftrightarrow\frac{x^2-\frac{7}{x^2}-\frac{9}{4}}{\sqrt{x^2-\frac{7}{x^2}}+\frac{3}{2}}+\frac{x-\frac{7}{x^2}-\frac{1}{4}}{\sqrt{x-\frac{7}{x^2}}+\frac{1}{2}}-\left(x-2\right)=0\)

\(\Leftrightarrow\frac{\frac{\left(4x^2+7\right)\left(x-2\right)\left(x+2\right)}{4x^2}}{\sqrt{x^2-\frac{7}{x^2}}+\frac{3}{2}}+\frac{\frac{\left(x-2\right)\left(4x^2+7x+14\right)}{4x^2}}{\sqrt{x-\frac{7}{x^2}}+\frac{1}{2}}-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{\frac{\left(4x^2+7\right)\left(x+2\right)}{4x^2}}{\sqrt{x^2-\frac{7}{x^2}}+\frac{3}{2}}+\frac{\frac{4x^2+7x+14}{4x^2}}{\sqrt{x-\frac{7}{x^2}}+\frac{1}{2}}-1\right)=0\)

Dễ thấy: \(\frac{\frac{\left(4x^2+7\right)\left(x+2\right)}{4x^2}}{\sqrt{x^2-\frac{7}{x^2}}+\frac{3}{2}}+\frac{\frac{4x^2+7x+14}{4x^2}}{\sqrt{x-\frac{7}{x^2}}+\frac{1}{2}}-1=0\) vô nghiệm

Nên \(x-2=0\Rightarrow x=2\)

Answer 2

[Nguyễn Duy Long]

thắng nguyễn chứng minh giùm hộ với... vì sao đống lăng nhăng đó lại vô nghiệm